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Here i am again newtons second law

by badman
Tags: newtons
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badman
#1
Jun6-05, 12:18 AM
P: 57
Consider another special case in which the inclined plane is vertical (\theta=\pi/2). In this case, for what value of m_1 would the acceleration of the two blocks be equal to zero?
Express your answer in terms of some or all of the variables m_2 and g.

a good point in the direction will help
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Pyrrhus
#2
Jun6-05, 12:28 AM
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Draw a Free Body Diagram for each box, and i recommend for the inclined box to have a system of coordinates at the angle theta with respect to the horizontal.
Sarahchichi
#3
Jun6-05, 07:11 AM
P: 12
The acceleration of the two blocks be equal to zero when the weight of Block 1 is equals to the tension of string. Then, there is no net force acting on Block 1, hence Block 2, the velocity is uniform or equals to zero.

badman
#4
Jun6-05, 09:43 AM
P: 57
Here i am again newtons second law

thanks guys. can you help me with a another problem.
Now imagine two points, Q and P, that divide the rope into segments L, M ,and R. View Figure The rope remains stationary. Assume that segment L exerts a force of magnitude F_LM on segment M. What is the magnitude F_RM of the force exerted by segment R on segment M?
Give your answer in terms of F_LM and constants such as g.

shouldnt the force acting on segment R be equal to the force segment M exerted on it but in opposite direction?

i inputed F_lm *G but it said i was missing a multiplicative factor?
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Doc Al
#5
Jun6-05, 09:51 AM
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Quote Quote by badman
i inputed F_lm *G but it said i was missing a multiplicative factor?
Why did you multiply by "G"?
badman
#6
Jun6-05, 11:03 AM
P: 57
whoops i was thinking of another answer to another problem i had. sorry bout that
badman
#7
Jun7-05, 09:58 AM
P: 57
nvm, i had another question but i already solved it
badman
#8
Jun7-05, 11:03 AM
P: 57
heres another i cant seem to right

The diagram below shows a block of mass m=2.0\; \rm kg on a frictionless horizontal surface, as seen from above. Three forces of magnitudes F_1 = 4.0\;{\rm N}, F_2 = 6.0\;{\rm N}, and F_3 = 8.0\;{\rm N} are applied to the block, initially at rest on the surface, at angles shown on the diagram. View Figure In this problem, you will determine the resultant (total) force vector from the combination of the three individual force vectors. All angles should be measured counterclockwise from the positive x axis (i.e., all angles are positive).

this is the question
Calculate the magnitude of the total resultant force \vec{F}_{\rm r} = \vec{F}_1+ \vec{F}_2 +\vec{F}_3 acting on the mass.
Express the magnitude of the resultant force in newtons to two significant figures.



i already know the formula and that i need to break this down into compenents. for each force vector.
the confusing thing is the angles, that i need.

this is basically how i set it up. Fx1= N*cos155
fy1=N*sin155
fx2 and fy2 i cant seem to get
fx3=N*cos325
fy3=Ncos325


are the angles right, i cant seem to figure out the angle for the vector F_2
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Doc Al
#9
Jun7-05, 12:18 PM
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Quote Quote by badman
this is basically how i set it up. Fx1= N*cos155
fy1=N*sin155
fx2 and fy2 i cant seem to get
fx3=N*cos325
fy3=Ncos325


are the angles right, i cant seem to figure out the angle for the vector F_2
The angles are not right. As marked in the diagram, the angle that each vector makes with the +x axis is:
F1 is at 25 degrees (not 155);
F2 is at 325 degrees;
F3 is at 180 degrees (it points in the -x direction, which is 180 degrees from the +x direction).
badman
#10
Jun7-05, 01:22 PM
P: 57
thanks doc.


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